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Impedance Triangle

In an R-L-C s...

Question

In an R-L-C series circuit shown in the figure, the readings of voltmeters $V_{1}$ and $V_{2}$ are 100 $V$ and 120 $V$ respectively. The source voltage is 130 $V$ . For this situation, mark out the correct statement(s).

Voltage across resistor, inductor and capacitor are

50 V, 50 \sqrt{3} V

and

(120 + 50 \sqrt{3}) V

respectively.

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Voltage across resistor, inductor and capacitor are

50 V, 50 \sqrt{3} V

and

120 - 50 \sqrt{3} V

respectively.

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Power factor of the circuit is

\frac{5}{13}

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The circuit is capacitive in nature.

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Solution

The correct options are
A Voltage across resistor, inductor and capacitor are $50 V, 50 \sqrt{3} V$ and $(120 + 50 \sqrt{3}) V$ respectively.
C Power factor of the circuit is $\frac{5}{13}$ .
D The circuit is capacitive in nature.
In R-L-C series circuit,
$V_{R}^{2} + V_{L}^{2} = 100^{2}$
$(V_{c} - V_{L} | = 120$

$130^{2} = V_{R}^{2} + {(V_{C} - V_{L})}_{C}^{2} \Rightarrow V_{R} = 50 V$
$\Rightarrow$ Voltage across inductor, $V_{L} = \sqrt{100^{2} - 50^{2}} = 50 \sqrt{3} V$
and voltage drop across capacitor,
$V_{C} = (120 + 50 \sqrt{3} V)$
Power factor, $cos ϕ = \frac{V_{R}}{V_{Z}} = \frac{50}{130} = \frac{5}{13}$
Since, $V_{C} > V_{L}$ , so circuit is capacitive.

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In an R-L-C series circuit shown in the figure, the readings of voltmeters V1 and V2 are 100 V and 120 V respectively. The source voltage is 130 V . For this situation, mark out the correct statement(s).

In an R-L-C series circuit shown in the figure, the readings of voltmeters $V_{1}$ and $V_{2}$ are 100 $V$ and 120 $V$ respectively. The source voltage is 130 $V$ . For this situation, mark out the correct statement(s).